To analyze the series $\sum_{n=0}^\infty \frac{3^n x^n}{(n+1)^2}$, we will determine whether it converges or diverges, find the radius and center of convergence, and discuss the behavior of the series at the endpoints.

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1. Identify the general term:

The general term of the series is: $a_n = \frac{3^n x^n}{(n+1)^2}.$

2. Apply the Ratio Test to find the radius of convergence:

The Ratio Test is useful for series involving powers of $x$. Define: $a_n = \frac{3^n x^n}{(n+1)^2}.$

Compute the ratio:

= \frac{3 x \cdot (n+1)^2}{(n+2)^2}.$$ Simplify: $$\frac{a_{n+1}}{a_n} = 3|x| \cdot \frac{(n+1)^2}{(n+2)^2}.$$ As $$n \to \infty$$, $$\frac{(n+1)^2}{(n+2)^2} \to 1$$. Thus: $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 3|x|.$$ For convergence, the Ratio Test requires: $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1.$$ Thus: $$3|x| < 1 \implies |x| < \frac{1}{3}.$$ The radius of convergence is: $$R = \frac{1}{3}.$$ ### 3. **Determine the center:** The center of convergence is at $$x = 0$$ (as the series is written in powers of $$x$$). ### 4. **Behavior at the endpoints ($$x = -\frac{1}{3}$$ and $$x = \frac{1}{3}$$):** We test the convergence of the series at $$x = -\frac{1}{3}$$ and $$x = \frac{1}{3}$$. #### Endpoint $$x = \frac{1}{3}$$: Substitute $$x = \frac{1}{3}$$ into the series: $$\sum_{n=0}^\infty \frac{(3 \cdot \frac{1}{3})^n}{(n+1)^2} = \sum_{n=0}^\infty \frac{1}{(n+1)^2}.$$ This is a $$p$$-series with $$p = 2 > 1$$, which converges. #### Endpoint $$x = -\frac{1}{3}$$: Substitute $$x = -\frac{1}{3}$$ into the series: $$\sum_{n=0}^\infty \frac{(3 \cdot -\frac{1}{3})^n}{(n+1)^2} = \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^2}.$$ This is an alternating series with terms $$\frac{1}{(n+1)^2}$$ that decrease monotonically to 0. By the Alternating Series Test, this series converges. ### 5. **Summary of Results:** - **Radius of convergence:** $$R = \frac{1}{3}$$. - **Center:** $$x = 0$$. - **Endpoint behavior:** - At $$x = \frac{1}{3}$$, the series converges. - At $$x = -\frac{1}{3}$$, the series converges. Thus, the interval of convergence is: $$\left[-\frac{1}{3}, \frac{1}{3}\right].$$ --- Do you need more details or have questions? ### Related Questions: 1. What is the Ratio Test, and how is it applied? 2. How can we generalize the radius of convergence for a power series? 3. What is the significance of a $$p$$-series, and why does it converge for $$p > 1$$? 4. Can the Alternating Series Test fail? Under what conditions? 5. What is the importance of endpoints in determining the interval of convergence? ### Tip: Always test the endpoints separately after determining the radius of convergence, as their behavior often differs from the rest of the interval.